Derivation

• December 1st 2007, 06:38 AM
Coach
Derivation

$h(x)=(x+1)^2(2x+1)^3$

Thank you!
• December 1st 2007, 07:10 AM
Krizalid
Quote:

Originally Posted by Coach
$h(x)=(x+1)^2(2x+1)^3$

Semi-nice way: product rule. (Do you know what the product rule is?)

Nasty way: expand the expression but this has a benefit, you'll only have to apply the power rule.
• December 1st 2007, 07:34 AM
Coach

I do know the product rule in case you are talking about the uv´+vu´ rule, where the accent indicates derivated(I am sorry, I am very bad at these terms, since the textbook we are using is in Finnish). However, whenever I try to apply this rule, I get the wrong answer. Here is what I did.

$(X+1)^3*2(2X-1)^1+ (2X-1)^2*3(X+1)^2$
• December 1st 2007, 07:50 AM
Krizalid
Yes, that states the product rule. (Don't worry about the terms, I'm chilean and I have to learn them anyway.)

$\Big[ {f(x) \cdot g(x)} \Big]' = f'(x) \cdot g(x) + f(x) \cdot g'(x).$

The function is $h(x)=(x+1)^2(2x+1)^3,$

So $h'(x)=2(x+1)\cdot(2x+1)^3+3(x+1)^2(2x+1)^2\cdot2.$

Can you take it from there?
• December 1st 2007, 07:58 AM
Coach
Yes. Thank you!

But where did you get the last 2 from?
• December 1st 2007, 08:03 AM
Krizalid
It's the chain rule; the derivative of $(2x+1)^3$ is $3(2x+1)^2\cdot(2x+1)'.$ That's why appears the $2.$